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Question

Assuming that water vapour is an ideal gas, the internal energy change (ΔU) when 1mol of water is vaporised at 1 bar pressure and 100oC will be:

(Given: Molar enthalpy of vapourisation of water at 1 bar and 373K=41 kJ.mol1 and R=8.3Jmol1 K1)

A
4.100 kJ mol1
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B
3.7904 kJ mol1
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C
37.904 kJ mol1
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D
41.00 kJ mol1
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Solution

The correct option is C 37.904 kJ mol1
The change :
H2O(l)H2O(g)

ΔH=ΔU+ΔngRT
"or" ΔU=ΔHΔngRT

Substituting the values, we get
ΔU=41.00kJ/mol1×8.3J/mol/K×373K×11000
=41.00kJmol13.096kJmol1
=37.904kJ/mol

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